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Magazine Chapter 7: Problem 85
Two linearly independent solutions \(-y_{1}\) and \(y_{2}\) -are given that satisfy the corresponding homogeneous equation. Use the method of variation of parameters to find a particular solution to the given nonhomogeneous equation. Assume \(x>0\) in each exercise. \(x^{2} y^{\prime \prime}-2 y=10 x^{2}-1\), \(y_{1}(x)=x^{2}, \quad y_{2}(x)=x^{-1}\)
Short Answer
Expert verified
The particular solution is:
\[y_p = \frac{1}{3} x^2 (1 - \ln |x|) + C.\]
Step by step solution
01
Write Down the Variation of Parameters Formulas
For a second-order linear differential equation with two solutions \(y_1(x)\) and \(y_2(x)\), the particular solution using variation of parameters is given by\[y_p = u_1(x) y_1(x) + u_2(x) y_2(x),\]where \(u_1(x)\) and \(u_2(x)\) are determined by solving the system:\[u_1'(x) y_1(x) + u_2'(x) y_2(x) = 0 \]and \[u_1'(x) y_1'(x) + u_2'(x) y_2'(x) = g(x),\]and \(g(x)\) is the nonhomogeneous part \(10x^2 - 1\).
02
Compute the Wronskian
The Wronskian, \(W\), for \(y_1(x) = x^2\) and \(y_2(x) = x^{-1}\) is computed as:\[W = \begin{vmatrix} x^2 & x^{-1} \ 2x & -x^{-2} \end{vmatrix} = x^2(-x^{-2}) - x^{-1}(2x) = -1 - 2x^1 = -1 - 2x^{-1}.\]
03
Determine u'_1(x) and u'_2(x)
Using the equations:\[u_1'(x) y_1(x) + u_2'(x) y_2(x) = 0 \] and\[u_1'(x) y_1'(x) + u_2'(x) y_2'(x) = g(x),\]we plug-in \(y_1\) and \(y_2\):\[u_1'(x) x^2 + u_2'(x) x^{-1} = 0\] \[u_1'(x) 2x + u_2'(x) (-x^{-2}) = 10x^2 - 1.\]
04
Substitute and Solve for u'_1(x) and u'_2(x)
To solve for \(u'_1(x)\) and \(u'_2(x)\) from:\[u_1'(x) x^2 + u_2'(x) x^{-1} = 0 \]\[u_1'(x) 2x + u_2'(x)(-x^{-2}) = 10x^2 - 1\]First take \(u_2'(x) = -x^3 u_1'(x)\) from the first equation and substitute it into the second equation:\[u_1'(x) 2x - x^3 u_1'(x) (-x^{-2}) = 10x^2 - 1\]=> \(2x u_1'(x) + x^1 u_1'(x) = 10x^2 - 1\)So, \[3x u_1'(x) = 10x^2 - 1\]=> \(u_1'(x) = \frac{10x^2 - 1}{3x}\).
05
Integrate to Get u_1(x) and u_2(x)
Integrate \(u_1'(x)\):\[u_1(x) = \int \frac{10x^2 - 1}{3x} \, dx\]This splits into:\[u_1(x) = \int \left(\frac{10x}{3} - \frac{1}{3x}\right) dx\]\[= \frac{10}{6} x^2 - \frac{1}{3} \ln |x| + C_1.\]For \(u_2(x)\), using \(u_2'(x) = -x^3 u_1'(x)\):\[u_2(x) = \int (-x^3 \frac{10x^2 - 1}{3x}) dx\]\[= \int (-\frac{10x^4 - x^2}{3}) dx\]\[= -\frac{10}{15} x^5 + \frac{1}{3} x^3 + C_2.\]
06
Substitute Back to Find Particular Solution
The particular solution becomes:\[y_p = u_1(x) y_1(x) + u_2(x) y_2(x) \]\[= \left(\frac{10}{6} x^2 - \frac{1}{3} \ln |x|\right)x^2 + \left(-\frac{10}{15} x^5 + \frac{1}{3} x^3\right)x^{-1} + C.\]Simplify:\[= \frac{10}{6} x^4 - \frac{1}{3} x^2 \ln |x| - \frac{10}{15} x^4 + \frac{1}{3} x^2 + C.\]Combine like terms to get:\[y_p = \frac{1}{3} x^2 - \frac{1}{3} x^2 \ln |x| + C.\]
07
Simplify the Result
Simplify the result of the particular solution:\[y_p = \frac{1}{3} x^2 (1 - \ln |x|) + C.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Wronskian
The Wronskian is a crucial concept when dealing with differential equations, especially in the method of variation of parameters. It is a determinant that helps us verify if a set of solutions is linearly independent. For a pair of solutions, such as \(y_1(x) = x^2\) and \(y_2(x) = x^{-1}\), the Wronskian \(W\) is calculated as:
- \( W = \begin{vmatrix} x^2 & x^{-1} \ 2x & -x^{-2} \end{vmatrix} = x^2(-x^{-2}) - x^{-1}(2x) \)
- Simplifying this, we get \( W = -1 - 2x^{-1} \)
Nonhomogeneous Differential Equation
A nonhomogeneous differential equation includes a term that does not solely involve the function and its derivatives. This equation can be expressed as a general form \(L(y) = g(x)\), where \(L(y)\) involves derivatives of \(y\), and \(g(x)\) is the nonhomogeneous part.
In the exercise you encountered, the equation is:
In the exercise you encountered, the equation is:
- \( x^2 y'' - 2y = 10x^2 - 1 \)
- Here, \(10x^2 - 1\) is the nonhomogeneous part \(g(x)\).
Particular Solution
The particular solution to a differential equation targets only the nonhomogeneous part \(g(x)\). It represents one possible solution that satisfies the entire equation, not just the associated homogeneous part.
In the method of variation of parameters, we derive a particular solution by using the functions \(y_1(x)\) and \(y_2(x)\), which solve the homogeneous part:
In the method of variation of parameters, we derive a particular solution by using the functions \(y_1(x)\) and \(y_2(x)\), which solve the homogeneous part:
- The particular solution \(y_p\) is structured as \(y_p = u_1(x) y_1(x) + u_2(x) y_2(x)\)
- Steps:
- Find derivatives \(u'_1(x)\) and \(u'_2(x)\) to satisfy the system:
\(u_1'(x) y_1(x) + u_2'(x) y_2(x) = 0\) - \(u_1'(x) y_1'(x) + u_2'(x) y_2'(x) = g(x)\)
- Find derivatives \(u'_1(x)\) and \(u'_2(x)\) to satisfy the system:
Method of Solution
The method of solution for finding a particular solution of a nonhomogeneous differential equation using variation of parameters involves several clear steps:
- Identify the Homogeneous Solution: Solve the related homogeneous equation to find solutions like \(y_1(x)\) and \(y_2(x)\), which are used in forming the Wronskian.
- Calculate the Wronskian: Ensure the solutions found are linearly independent by verifying their Wronskian is non-zero.
- Set Up and Solve the System: Form the system of equations based on the derivatives \(u_1'(x)\) and \(u_2'(x)\), then solve these to find expressions for each.
- Integrate to Find \(u_1(x)\) and \(u_2(x)\): Integrate the expressions \(u_1'(x)\) and \(u_2'(x)\) to derive the functions.
- Construct the Particular Solution: Combine using the formula \(y_p = u_1(x) y_1(x) + u_2(x) y_2(x)\).
